If by CLT $\sqrt{N}\epsilon_N \rightarrow v$, where $v$ is $N(0,1)$, does that imply that $\epsilon_N \rightarrow 1/\sqrt{N}v$?

Question

If, by the central limit theorem, $\sqrt{N}\epsilon_N \rightarrow v$ in distribution as $N \rightarrow \infty$, where $v$ is $N(0,1)$, does that imply that $\epsilon_N \rightarrow 1/\sqrt{N}v$?

In one of my text books it is stated that $\sqrt{N}\epsilon_N \rightarrow v$, but I'm interested in the distribution of $\epsilon_N$.

It seems a bit strange to consider $\epsilon_N \rightarrow 1/\sqrt{N}v$ as a limit value since $N$ is by definition already infinitely large, so maybe $\sqrt{N}\epsilon_N \rightarrow v \implies \epsilon_N \rightarrow 0$?

Answer 1

If $\sqrt {N} \epsilon_N \to v$ in distribution then $\epsilon_n \to 0$ in distribution and in probability.

[Proof: Choose $T$ such that $P(v >T) <\epsilon$. Then $$\lim \sup P(\epsilon_N >t)$$ $$ =\lim \sup P( \sqrt {N} \epsilon_N >t\sqrt {N})\leq \lim \sup P( \sqrt {n} \epsilon_N >T)$$ $$ =P(v>T) <\epsilon$$ for $N$ sufficiently large].